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IPMAT Indore 2019 (MCQ) PYQs

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IPMAT Indore 2019

Arithmetic
>
Simple & Compound Interest

Easy

If the compound interest earned on a certain sum for 2 years is twice the amount of simple interest for 2 years, then the rate of interest per annum is _______ percent

✅ Correct Option: 1
We need to find the rate of interest when compound interest for 2 years is twice the simple interest for 2 years.
Let's define our variables: \newline - Principal = PP \newline - Rate = RR% per annum \newline - Time = 22 years
Simple Interest formula:
SI=P×R×T100SI = \dfrac{P \times R \times T}{100}
For 2 years:
SI=P×R×2100=2PR100SI = \dfrac{P \times R \times 2}{100} = \dfrac{2PR}{100}
Compound Interest formula:
CI=P[(1+R100)T−1]CI = P\left[\left(1 + \dfrac{R}{100}\right)^T - 1\right]
For 2 years:
CI=P[(1+R100)2−1]CI = P\left[\left(1 + \dfrac{R}{100}\right)^2 - 1\right]
Expanding:
CI=P[1+2R100+R210000−1]CI = P\left[1 + \dfrac{2R}{100} + \dfrac{R^2}{10000} - 1\right]
CI=P[2R100+R210000]CI = P\left[\dfrac{2R}{100} + \dfrac{R^2}{10000}\right]
CI=2PR100+PR210000CI = \dfrac{2PR}{100} + \dfrac{PR^2}{10000}
Given condition: CI=2×SICI = 2 \times SI
2PR100+PR210000=2×2PR100\dfrac{2PR}{100} + \dfrac{PR^2}{10000} = 2 \times \dfrac{2PR}{100}
2PR100+PR210000=4PR100\dfrac{2PR}{100} + \dfrac{PR^2}{10000} = \dfrac{4PR}{100}
Subtracting 2PR100\dfrac{2PR}{100} from both sides:
PR210000=4PR100−2PR100\dfrac{PR^2}{10000} = \dfrac{4PR}{100} - \dfrac{2PR}{100}
PR210000=2PR100\dfrac{PR^2}{10000} = \dfrac{2PR}{100}
Dividing both sides by PP:
R210000=2R100\dfrac{R^2}{10000} = \dfrac{2R}{100}
Multiplying both sides by 100100:
R2100=2R\dfrac{R^2}{100} = 2R
Dividing both sides by RR (assuming R≠0R \neq 0):
R100=2\dfrac{R}{100} = 2
Therefore:
R=200R = 200
The rate of interest per annum is 200200 percent.

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